[proofplan] We change variables from canonical coefficient vectors to Euclidean unit vectors by whitening the two covariance blocks. Under this transformation, the canonical covariance objective becomes the [bilinear form](/page/Bilinear%20Form) $\alpha^{\top}C\beta$ on unit spheres. The successive constrained maxima of this bilinear form are exactly the singular values of $C$, and squaring those singular values gives the eigenvalues of $C C^{\top}$ or $C^{\top}C$. Finally, multiplying out $C C^{\top}$ identifies the eigenvalue matrix with $M$. [/proofplan] [step:Whiten the covariance constraints into Euclidean unit constraints] Because $\Sigma_{11}$ and $\Sigma_{22}$ are symmetric positive definite, their positive definite square roots $\Sigma_{11}^{1/2}$ and $\Sigma_{22}^{1/2}$ exist and are invertible. Define the change of variables \begin{align*} \alpha := \Sigma_{11}^{1/2}a \in \mathbb{R}^p, \qquad \beta := \Sigma_{22}^{1/2}b \in \mathbb{R}^q. \end{align*} Equivalently, \begin{align*} a = \Sigma_{11}^{-1/2}\alpha, \qquad b = \Sigma_{22}^{-1/2}\beta. \end{align*} Then \begin{align*} a^{\top}\Sigma_{11}a &= \alpha^{\top}\Sigma_{11}^{-1/2}\Sigma_{11}\Sigma_{11}^{-1/2}\alpha = |\alpha|^2, \\ b^{\top}\Sigma_{22}b &= \beta^{\top}\Sigma_{22}^{-1/2}\Sigma_{22}\Sigma_{22}^{-1/2}\beta = |\beta|^2. \end{align*} Thus the constraints $a^{\top}\Sigma_{11}a=1$ and $b^{\top}\Sigma_{22}b=1$ are exactly the Euclidean constraints $|\alpha|=1$ and $|\beta|=1$. Moreover, the objective transforms as \begin{align*} a^{\top}\Sigma_{12}b &= \alpha^{\top}\Sigma_{11}^{-1/2}\Sigma_{12}\Sigma_{22}^{-1/2}\beta = \alpha^{\top}C\beta. \end{align*} The canonical orthogonality constraints transform in the same way: \begin{align*} a^{\top}\Sigma_{11}a_j = \alpha^{\top}\alpha_j, \qquad b^{\top}\Sigma_{22}b_j = \beta^{\top}\beta_j, \end{align*} where $\alpha_j := \Sigma_{11}^{1/2}a_j$ and $\beta_j := \Sigma_{22}^{1/2}b_j$. [/step] [step:Identify the successive maxima with the singular values of $C$] Let $r := \operatorname{rank}(C)$. By the [Singular Value Decomposition](/theorems/???), choose orthonormal bases \begin{align*} u_1,\dots,u_p \in \mathbb{R}^p, \qquad v_1,\dots,v_q \in \mathbb{R}^q \end{align*} and numbers \begin{align*} \sigma_1 \geq \sigma_2 \geq \cdots \geq \sigma_s \geq 0 \end{align*} such that $\sigma_i>0$ for $1 \leq i \leq r$, $\sigma_i=0$ for $r<i\leq s$, and the full diagonal action is \begin{align*} C v_i = \sigma_i u_i, \qquad C^{\top}u_i = \sigma_i v_i \end{align*} for $1 \leq i \leq s$. If $q>s$, then $C v_j=0$ for $s<j\leq q$; if $p>s$, then $C^{\top}u_i=0$ for $s<i\leq p$. For unit vectors $\alpha \in \mathbb{R}^p$ and $\beta \in \mathbb{R}^q$, write \begin{align*} \alpha = \sum_{i=1}^{p} x_i u_i, \qquad \beta = \sum_{j=1}^{q} y_j v_j, \end{align*} where $\sum_{i=1}^{p}x_i^2=1$ and $\sum_{j=1}^{q}y_j^2=1$. Then \begin{align*} \alpha^{\top}C\beta = \sum_{i=1}^{s}\sigma_i x_i y_i. \end{align*} By the [Cauchy-Schwarz Inequality](/theorems/???) applied in $\mathbb{R}^s$ to the vectors $(\sigma_i x_i)_{i=1}^s$ and $(y_i)_{i=1}^s$, \begin{align*} \alpha^{\top}C\beta \leq \left(\sum_{i=1}^{s}\sigma_i^2 x_i^2\right)^{1/2} \left(\sum_{i=1}^{s}y_i^2\right)^{1/2} \leq \sigma_1. \end{align*} Equality is attained at $\alpha=u_1$ and $\beta=v_1$, so $\rho_1=\sigma_1$. For the $k$-th constrained maximisation, we choose the already constructed canonical directions to be the SVD pairs $(u_i,v_i)$ for $1 \leq i<k$. This choice is legitimate because canonical directions are only determined up to an orthonormal change of basis inside a repeated singular-value subspace, while the canonical correlations are the ordered constrained maximum values. With this SVD choice, the transformed orthogonality conditions are \begin{align*} \alpha \perp u_1,\dots,u_{k-1}, \qquad \beta \perp v_1,\dots,v_{k-1}. \end{align*} Thus $x_i=y_i=0$ for $1 \leq i<k$, and the same computation gives \begin{align*} \alpha^{\top}C\beta = \sum_{i=k}^{s}\sigma_i x_i y_i \leq \sigma_k. \end{align*} Equality is attained at $\alpha=u_k$ and $\beta=v_k$. If a different orthonormal basis is chosen inside a repeated singular-value subspace, the constraints remove the same number of dimensions from that subspace, and the displayed SVD coordinate estimate on the remaining orthogonal complement gives the same next ordered value. Therefore \begin{align*} \rho_k=\sigma_k \end{align*} for every $1 \leq k \leq s$. [/step] [step:Square the singular values to obtain the eigenvalue problem] For $1 \leq i \leq s$, the paired singular-vector relations $C v_i=\sigma_i u_i$ and $C^{\top}u_i=\sigma_i v_i$ give \begin{align*} C C^{\top}u_i &= C(\sigma_i v_i) = \sigma_i C v_i = \sigma_i^2 u_i, \\ C^{\top}C v_i &= C^{\top}(\sigma_i u_i) = \sigma_i C^{\top}u_i = \sigma_i^2 v_i. \end{align*} Hence the squared singular values $\sigma_i^2$ are the eigenvalues of $C C^{\top}$ on the left canonical space and of $C^{\top}C$ on the right canonical space, with zeros included up to $s=\min(p,q)$. Since $\rho_i=\sigma_i$ for $1 \leq i \leq s$, the squared canonical correlations satisfy \begin{align*} \rho_i^2=\sigma_i^2. \end{align*} Thus the squared canonical correlations are exactly the eigenvalues of the corresponding smaller-dimensional singular-value eigenproblem, counted with multiplicity. [/step] [step:Compute $C C^{\top}$ and identify it with $M$] Using the definition \begin{align*} C=\Sigma_{11}^{-1/2}\Sigma_{12}\Sigma_{22}^{-1/2}, \end{align*} and using $\Sigma_{21}=\Sigma_{12}^{\top}$ together with the symmetry of $\Sigma_{22}^{-1/2}$, we compute \begin{align*} C C^{\top} &= \Sigma_{11}^{-1/2}\Sigma_{12}\Sigma_{22}^{-1/2} \left(\Sigma_{11}^{-1/2}\Sigma_{12}\Sigma_{22}^{-1/2}\right)^{\top} \\ &= \Sigma_{11}^{-1/2}\Sigma_{12}\Sigma_{22}^{-1/2} \Sigma_{22}^{-1/2}\Sigma_{21}\Sigma_{11}^{-1/2} \\ &= \Sigma_{11}^{-1/2}\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\Sigma_{11}^{-1/2} \\ &= M. \end{align*} Therefore, when $p \leq q$, the numbers $\rho_1^2,\dots,\rho_s^2$ are precisely the eigenvalues of $M=C C^{\top}$ listed in non-increasing order and counted with multiplicity. When $q \leq p$, the same squared values are equivalently the eigenvalues of $C^{\top}C$ on the smaller right canonical space, with the remaining eigenvalues of $M$ equal to zero if $p>q$. This proves both the eigenvalue formulation and the singular-value formulation. [/step]